LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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The minimum-degree condition $δ (G) \geq n / 2$ implies Ore's degree-sum condition

Statement

Let $G$ be a finite simple graph on $n \geq 1$ vertices. If $δ (G) \geq n / 2$ , then every nonadjacent pair $u, v$ satisfies $\deg (u) + \deg (v) \geq n$ .

Facts & Assumptions

Given: A nonempty finite simple graph $G$ on $n$ vertices with $δ (G) \geq n / 2$ .

[F1]

The minimum degree satisfies $δ (G) \leq \deg (w)$ for every vertex $w$ (Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

[F2]

The order $n$ is the finite cardinality of the vertex set (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1

For any nonadjacent vertices $u, v$ , [F1] and the hypothesis give $\deg (u) + \deg (v) \geq 2 δ (G) \geq n$ .

givenF1F2algebra

2.1

Since the pair was arbitrary, Ore's degree-sum condition holds for every nonadjacent pair.

step 1.1∎

Depends on

Used by

Dirac's theorem: every n-vertex graph with n≥3 and δ(G)≥ n/2 is Hamiltonian Corollary

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Applied Combinatorics, Eulerian and Hamiltonian Graphs (standard reference, not scraped)

The minimum-degree condition δ(G)≥n/2δ(G)≥n/2 implies Ore's degree-sum condition